Question

Two blocks of masses in the ratio \( m:n \) are connected by a light inextensible string passing over a frictionless fixed pulley. If the system of the blocks is released from rest, then the acceleration of the centre of mass of the system of the blocks is (g = acceleration due to gravity)

• A. \( \left(\frac{m+n}{m-n}\right)^2 g \)
• B. \( \left(\frac{m-n}{m+n}\right)^2 g \)
• C. \( \left(\frac{m+n}{m-n}\right) g \)
• D. \( \left(\frac{m-n}{m+n}\right) g \)

Hint:

Acceleration of block in Atwood machine: \( a = \frac{\Delta m}{\sum m} g \). Acceleration of Center of Mass: \( a_{cm} = \left( \frac{\Delta m}{\sum m} \right) a = \left( \frac{\Delta m}{\sum m} \right)^2 g \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We have an Atwood machine with masses M1 and M2 where M1 / M2 = m / n.
We need to find the acceleration of the center of mass acm.
The individual acceleration of the blocks is a.
acm = (M1a1 + M2a2) / (M1 + M2)

Step 2: Key Formula or Approach:

1. Acceleration of blocks:
a = |M1 - M2| / (M1 + M2) × g

2. Acceleration of center of mass:
Since one block moves up and the other moves down, we take
a1 = a and a2 = -a (or vice versa).

So,
acm = (M1(a) + M2(-a)) / (M1 + M2)
= a(M1 - M2) / (M1 + M2)

Step 3: Detailed Explanation:

Let the masses be proportional to m and n.
The proportionality constant cancels in ratios, so we can directly use m and n.

Acceleration of the blocks:
a = (m - n) / (m + n) × g

Assume m > n, so the mass m moves downward and the mass n moves upward.

Now, acceleration of the center of mass is:
acm = (m·a + n·(-a)) / (m + n)

= a(m - n) / (m + n)

Substitute the value of a:
acm = [(m - n) / (m + n) × g] × [(m - n) / (m + n)]

acm = ((m - n) / (m + n))2 × g

Step 4: Final Answer:

The acceleration of the centre of mass is
((m - n) / (m + n))2 × g.